1000 Solved Problems in Modern Physics

(Tina Meador) #1

564 10 Particle Physics – II


Therefore,σ 1 =C|M 3 |^2
whereC=constant
For reaction (2) we have the mixture ofI= 1 /2 andI= 3 /2 states.
Refering to the table for 3/ 2 × 1 /2 C.G. coefficients we can write

|ψi〉=


∣ψf〉=


1

3




∣φ

(

3

2

,−

1

2

)〉



2

3




∣φ

(

1

2

,−

1

2

)〉

Therefore, σ 2 =C


ψf|H 1 +H 3 |ψi

〉 2

=C





1

3

M 3 +

2

3

M 1





2

For the reaction (3) we have

|ψi〉=


1

3




∣φ

(

3

2

,−

1

2

)〉



2

3




∣φ

(

1

2

,−

1

2

)〉


∣ψf


=


2

3




∣φ

(

3

2

,−

1

2

)〉

+


1

3




∣φ

(

1

2

,−

1

2

)〉

Therefore,σ 3 =C





2
9 M^3 −


2
9 M^1




2

The ratio of cross-sections are

σ 1 :σ 2 :σ 3 =|M 3 |^2 :

1

9

|M 3 + 2 M 1 |^2 :

2

9

|M 3 −M 1 |^2

Ifa 1 / 2 <<a 3 / 2 ,M 1 <<M 3 ,σ 1 :σ 2 :σ 3 =9:1:2
And ifa 1 / 2 >>a 3 / 2 ,M 1 >>M 3 ,σ 1 :σ 2 :σ 3 =0:2:1

10.19 From the knownπ−Nscattering cross-sections in terms of the amplitudes
a 3 / 2 anda 1 / 2 obtained in Problem 10.18 we can construct a diagram for the
amplitudes in the complex plane, Fig. 10.3


Fig. 10.3Diagram of
amplitudes



σ+=|a 3 |

σ−=

1

3

|a 3 + 2 a 1 |

2 σ^0 =

2

3

|a 3 −a 1 |

where for brevity we have writtena 1 fora 1 / 2 anda 3 fora 3 / 2
From the triangle the required inequality follows from the fact that the sum
of two sides is equal to or greater than the third side.
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